Question

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Answer 1

Answer:

Let Sn = Sum of 'n' terms, then,

Sn = a + (a+d) + (a+2d) + (a+3d) +...+a+(n-1)d

Let | = last term, then,

Sn = a + (a+d) + (a+2d) + (a+3d) +...+ (1-3d) + (1 -2d) + (1 - d) + 1

reversing the order this gives:

Sn=1+ (1-d) + (1-2d) + (1-3d) +...+ (a+3d) + (a +2d) + (a + d) + a

Adding these last two equations gives,

2*Sna+l+ (a+l) + (a+l) + (a+l) +...+ (a+l) + (a+l) + (a+l) +a+l

2*Sn = (a+l) to n terms

2*Sn = = (a+l)*n

Sn = n/2*(a+l)

but I = a+(n-1)d, so substituting this gives:

Sn = n/2 (2a+(n-1)d)